AIME 2000 II · 第 9 题

Solution

Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$, we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$.

There are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\theta} = \cos \theta + i\sin \theta$ and $\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta$. We have $z+\frac 1z = 2\cos \theta = 2\cos 3^\circ$ and $\theta = 3^\circ$. Alternatively, we could let $z = a + bi$ and solve to get $z=\cos 3^\circ + i\sin 3^\circ$.

Using De Moivre's Theorem we have $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ$, $6000 = 16(360) + 240$, so $z^{2000} = \cos 240^\circ + i\sin 240^\circ$.

We want $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$.

Finally, the least integer greater than $-1$ is $\boxed{000}$.

Solution 2

Let $z=re^{i\theta}$. Notice that we have $2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.$

$r$ must be $1$ (or if you take the magnitude would not be the same). Therefore, $z=e^{i\frac{\pi}{\theta}}$ and plugging into the desired expression, we get $e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1$. Therefore, the least integer greater is $\boxed{000}.$

~solution by williamgolly

Solution 3 Intuitive

For this solution, we assume that $z^{2000} + 1/z^{2000}$ and $z^{2048} + 1/z^{2048}$ has the same least integer greater than their solution.

We have $z + 1/z = 2\cos 3$. Since $\cos 3<1$, $2\cos 3<2$. If we square the equation $z + 1/z = 2\cos 3$, we get $z^2 + 2 + 1/(z^2) = 4\cos^2 3$, or $z^2 + 1/(z^2) = 4\cos^2 3 - 2$. $4\cos^2 3 - 2$ is is less than $2$, since $4\cos^2 3$ is less than $4$. If we square the equation again, we get $z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2$.

Since $4\cos^2 3 - 2$ is less than 2, $(4\cos^2 3 - 2)^2$ is less than 4, and $(4\cos^2 3 - 2)^2 -2$ is less than 2. However $(4\cos^2 3 - 2)^2 -2$ is also less than $4\cos^2 3 - 2$. we can see that every time we square the equation, the right-hand side gets smaller and into the negatives. Since the smallest integer that is allowed as an answer is 0, the smallest integer greater is $\boxed{000}.$

~ PaperMath ~megaboy6679

Solution 4

First, let $z = a+bi$ where $a$ and $b$ are real numbers. We now have that

$$a+bi+\frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}$$ given the conditions of the problem. Equating imaginary coefficients, we have that

$$b \left( 1 - \frac{1}{a^2+b^2}\right) = 0$$ giving us that either $b=0$ or $|z| = 1$. Let's consider the latter case for now.

We now know that $a^2+b^2=1$, so when we equate real coefficients we have that $2a = 2 \cos{3^{\circ}}$, therefore $a = \cos{3^{\circ}}$. So, $b = \sin{3^{\circ}}$ and then we can write $z = \text{cis}(3)^{\circ}$.

By De Moivre's Theorem,

$$z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}$$ . The imaginary parts cancel, leaving us with $2 \cos{6000^{\circ}}$, which is $240 \pmod{360}$. Therefore, it is $-1$, and our answer is $\boxed{000}$.

Now, if $b=0$ then we have that $a+\frac{1}{a} = 2 \cos{3^{\circ}}$. Therefore, $a$ is not violating our conditions set above.